TITLE: What is $\varlimsup \frac{\omega(n)}{\log n}$?
QUESTION [5 upvotes]: $\omega(n)$ is the number of distinct prime divisors of $n$. How to figure out?

 $$\varlimsup_{n\to\infty} \frac{\omega(n)}{\log n}$$

or $ \dfrac{\omega(n)}{\log n}$ is convergent, so  $\lim\limits_{n\to\infty} \frac{\omega(n)}{\log n}=0 $?


It is obvious that $2^{\omega(n)}\leq n$, or
$$\omega(n)\leq \frac{\log n}{\log2} $$
so $\varlimsup_{n\to\infty} \frac{\omega(n)}{\log n}\leq\frac1{\log 2}$
Continue to move forward,   $n\geq7$, then  $\omega(n)\leq \log n $. so 
$$\varlimsup_{n\to\infty} \frac{\omega(n)}{\log n}\leq1$$
P.s.  I think about this problem after reading To show: $\omega(n)\ne\pi(n) , \forall n>2$

REPLY [3 votes]: The largest value of this for each $\omega(n)$ is the primorial with that number of prime factors. With $\theta(x)$ Chebyshev's first function, you are asking about
$$ \limsup \frac{\pi(x)}{\theta(x)} \approx \frac{x}{\log x \; \; x} \rightarrow 0. $$